Define $$f(x)=\frac{\cos^2(\pi x)}{2+\cos(x)}$$ We know that $f(x)$ is not periodic. Is there any way to write $f(x)$ as the sum of two periodic functions. That is, find periodic functions $f_1(x)$ and $f_2(x)$ such that $f(x)=f_1(x)+f_2(x)$.
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No. Suppose $f=f_1+f_2$, where $f_1$ has period $p_1$ and $f_2$ has period $p_2$. Then $$f(x+p_1)=f_1(x+p_1)+f_2(x+p_1) = f_1(x)+f_2(x+p_1),$$ and $$f(x+p_1+p_2)=f_1(x+p_2)+f_2(x+p_1);$$ similarly, $$f(x+p_2)=f_1(x+p_2)+f_2(x),$$ for all $x$. Combining these, we see that for all $x$, we have $$ f(x+ p_1+p_2)-f(x+p_2)-f(x+p_1)+f(x)=0.\tag{*}$$
After clearing fractions in (*) you get a non-trivial trigonometric polynomial that vanishes for all values of $x$.
answered Apr 3, 2019 at 20:40
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1$\begingroup$ Thanks for your response. Could you please add more details. I did not get why the sum that you wrote is equal to 0. $\endgroup$– ArthurApr 3, 2019 at 20:44
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1$\begingroup$ @Arthur The explanations about the first identity are given after it. $\endgroup$ Apr 3, 2019 at 21:28
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1$\begingroup$ @JeanMarie I have rewritten my answer to make the argument a little more direct. In fairness to Arthur, I had no explanation of my (*) when he posted his comment; my edits were in response to his comment. $\endgroup$ Apr 3, 2019 at 21:41
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1$\begingroup$ I just wrote an explanation that uses Fourier Transform. @arthur : I am conscious that this explanation must be very difficult to follow. But maybe it can have a triggering effect on you as it had, long years ago, on me "I have to understand this fascinating domain" (called harmonic analysis) :) $\endgroup$ Apr 3, 2019 at 22:34
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